Transcript Giancoli, PHYSICS,6/E

Chapter 4
Motion and Force
Dynamics
© 2006 Giancoli, PHYSICS,6/E © 2004.
Electronically reproduced by permission of
Pearson Education, Inc., Upper Saddle River, New
Jersey
Module 8
Newtons’s Laws of Motion
Giancoli, Sec 4-1 4-5
Giancoli, PHYSICS,6/E © 2004. Electronically
reproduced by permission of Pearson Education, Inc.,
Upper Saddle River, New Jersey
Newton’s First Law of Motion
Aristotle said that force was necessary to make an object move
with constant velocity.
Example: Pull a box across the table. Must pull to keep it
going.
(Aristotle was fixated on friction)
Module 8 - 1
Newton’s First Law of Motion
Newton: asked what would happen if friction could be
eliminated.
“Every body continues in its state of rest or uniform speed in a
straight line unless acted on by a nonzero net force.”
Module 8 - 2
Newton’s Second Law of Motion
The acceleration of an object is directly proportional to the
net force acting on it and is inversely proportional to its
mass. The direction of the acceleration is in the direction of
the net force acting on the object

a

F
m
Notice that mass is a measure of an object’s resistance to
acceleration. We usually write the equation as


F  ma
Module 8 - 3
Notes on Forces


F

m
a

Units:
FORCE:
newton (N) 1 N = kg · m /s2
MASS:
kilogram (kg)
Since this is a vector equation, it can be written in
component form:
F
 Fy
 may
F
 maz
x
z
Module 8 - 4
 max
Example 4-1 (9) A 0.140-kg baseball traveling 35.0 m/s strikes the
catchers mitt, which, in bringing the ball to rest recoils backward 11.0
cm. What was the average force applied by the glove on the ball?
x   0.110 m ( to the right )
First find a:
v0   35.0 m
v0
s
v  v  2a( x  x )
2
2
0
0


2
m
v v
0  35.0
a
s

2( x  x )
2(0.110 m)
2
2
0
 5.57  10 m
3
0


 F  ma
s
2
F  (0.140kg )(5.57  10 m )  780 N (to the left)
s
Module 8 - 5
Force is in opposite direction to velocity.
3
2
Newton’s Third Law of Motion
Law of Action - Reaction
Whenever one object exerts a force on a second object, the
second object exerts an equal and opposite force on the first.
Examples:
skater leans on wall
wall exerts an equal but opposite
force on skater
Earth exerts a force on moon
moon exerts equal but opposite
force on earth
Module 8 - 6
Newton’s Third Law of Motion
Whenever one object exerts a force on a second object, the
second object exerts an equal and opposite force on the first.
Is important to realize that these forces act on different things
and thus they don’t cancel.
Module 8 - 7
Module 9
Applications of Newton’s Laws
Giancoli, Sec 4-6  4-7
Giancoli, PHYSICS,6/E © 2004. Electronically
reproduced by permission of Pearson Education, Inc.,
Upper Saddle River, New Jersey
Weight
Galileo told us that all objects experience an acceleration due
to gravity of g and Newton gave us F = ma. We can
combine these to ideas to realize that the weight of an object
is the force of attraction that the earth exerts on objects and it
can be written


FG  m g
It should be noted that weight is a force and thus the proper
units are newtons (N) or pounds. It is technically incorrect to
say that something weights 2.0 kg because that is the mass of
the object.
Module 9 - 1
Application: Ropes
Tension: when the man pulls on the rope, the tension in the
rope transmits the force to the box. The tension is 100 N
which exerts a 100 N force upward on the box and a 100 N
downward force on the hand.
98 N
m
 10 kg
m
9.8 2

s
FP
F
y

mg
Fp
 may
mg
 ma
100 N  98 N  10 kg a
a  0.20 m
Module 9 - 2
s2
Steps in Solving Problems
1. Draw free-body diagram for every object that is ”free”
2. Select coordinate system such that one of the axis is
along the direction of acceleration
3. Write out the equations of motion for the x and y
coordinate:
F
x
F
y
 max
 may
4. Step 2 should guarantee that the sum of the forces in all
but one direction equals zero.
5. Solve the equations simultaneously
Module 9 - 3
Application: Normal Force
When a box rests on a table, the
table must exert enough upward
force to support the box, otherwise,
the table will collapse. This upward
force is called the normal force FN
because it is normal to the surface.
Module 9 - 4
When we push
down with a force of
40 N the normal
force will increase
by 40 N.
Free-Body Diagrams
•Essential part of solution
•Vital tool to understand problem
•Forces are the only vectors on free-body diagrams
•If there are two objects, each of them will have a
free-body diagram
•If there are two objects, label each mass properly: m1
and m2
•Select a coordinate system such that the acceleration
direction is along one axis
•Then apply Newton’s Second Law (two equations
for each object):
Module 9 - 5
F
x
 max
F
y
 may
Example 4-2 A 65-kg woman ascends in an
elevator that briefly accelerates at 1.0 m/s2
upward when leaving a floor. She stands on a
scale that reads in N.
F ma
FN  m g  m a

FN
FN  m a  m g  m ( a  g )
FN  65 kg (1.0 m
s
2
 9.8 m
s
2
)  700 N
When acceleration is zero, the scale reads her weight:
FN  (65 kg) ( 9.8 m
Module 9 - 6
s
2
)  640 N

mg
Application: Ropes and Pulleys

FT 1

m1 g

FT 2

m2 g
• A pulley changes the direction of the tension in the rope.
• If the pulley is frictionless and massless then the
tension in the left rope is the same as the right


FT 1  FT 2
Module 9 - 7
Example 4-3 Two masses hang from a massless, frictionless pulley as shown. Draw
free-body diagram for each of the masses. Derive a formula for the acceleration of
the masses. Assume m1 = 0.250 kg and m2 = 0.200 kg.

FT

FT

m2 g

m1 g
F  m a
F m g ma
y
T
2
2
2
F mgma
T
2
DOWN   for m1 

UP   for m2

so a is the same for both
F  ma
m g  F  ma
y
1
1
T
1
2
m g  m g  m a  ma
1
Module 9 - 8
2
2
m m 
 g
a  
m m 
1
2
1
2
1
m2 m1
Example 4-3 Two masses hang from a massless, frictionless pulley as shown.
Draw free-body diagram for each of the masses. Calculate the acceleration of the
masses and the tension. Assume m1 =0.250 kg and m2 = 0.200 kg.

FT

FT

m2 g

m1 g
m2 m1
m m 
 g
a  
m m 
1
2
1
2
 0.250 kg  0.200 kg  m
9.8 2
a  
s
 0.250kg  0.200 kg 
 1.09 m
s
2
F  m g  m a  m2 ( g  a )
T
Module 8 - 1
2
2
FT  ( 0.200 kg ) ( 9.80 m
s
2
 1.09 m
s
2
)  2.18 N
Comments on Example 4-12 in Book
Treat as a single mass:
 F  m  m a
FP  mA  mB  a
x
A
B
FP
40.0 N
a

1.82 m
s
(m A  mB ) (10.0 kg  12.0 kg)
•Each box has the same acceleration
aA = aB = 1.82 m/ s2
• FT is not equal to FP
• FP = 40 N
• FT = (12 kg) ( 1.82 m/s2 ) = 22 N
Module 9 - 9
Module 10
Friction and Inclined Planes
Giancoli, Sec 4-8 4-9
Giancoli, PHYSICS,6/E © 2004. Electronically
reproduced by permission of Pearson Education, Inc.,
Upper Saddle River, New Jersey
Kinetic Friction
•Friction results when two
surfaces slide across each other
because even the smoothest
surfaces have some roughness.
Kinetic Friction results when a body
slides across a surface. It is proportional
to the normal force between the surfaces:
Fk   k FN
where k is a unit-less number called the coefficient
of kinetic friction.
Module 10 - 1
Static Friction
If we gradually increase the applied force
by adding water, the static friction force
matches it until the object starts to move.
Once it is sliding, the friction is kinetic and
is constant.
Static Friction: arises as a result of an
external force even when the body is not
yet moving:
Fs   s FN
Module 10 - 2
Example 4-4 From the data in the graph, determine µk and µk.

FN
 
FT  Ffr

mg
FN  mg  28 kg 9.80 m s   270 N
Fk   k FN  30 N
Fk
30 N

 0.11
k 
FN 270 N
Fs   s FN
Just before the box starts to move
Fs   s FN  39 N
k 
Fk
FN

39 N
 0.14
270 N
Module 10 - 3
Module 10 - 4
y

F fr
Inclined Planes

FN

x
mg

•An inclined plane exerts a normal force FN which is
perpendicular to the surface.
•There may also be a frictional force which opposes the motion.
•It should also be noted that the angle between the weight and the
normal  is the same as the angle of the incline .
Module 10 - 5
y
Example 4-5 A block of wood rests on
a wooden board. Derive the equations
of motion

F fr

FN

F  0
y
F  mg cos  0
N
FN  mg cos
F fr   FN   mg cos 
Module 10 - 6
x
mg
F
x

 ma
mg sin   F fr  ma
mg sin    m g cos   ma
y
Example 4-5 A block of wood rests on
a wooden board. One end of the board
is raised until the block starts to slip.
Determine the coefficient of static
friction if θ = 250 when it starts to slip.

F fr

FN

mg sin    m g cos   ma
At the point where it starts to slip a ≈ 0 and  = s.
g sin   s g cos  0
sin 
 
 tan
cos
 s  tan 250  0.47
s
Module 10 - 7
x
mg

Example 4- 6 One 2.80 kg paint bucket (m1) is hanging by a massless cord
from a 3.50 kg paint bucket (m2), also hanging by a massless cord. If the two
buckets are pulled upward with an acceleration of 0.700 m/s2 by the upper cord,
calculate the tension in each cord.
F  ma
F  m g  ma
1
T1

FT 2
m2g
1
F  m ( g  a)
T1
1
FT 1  ( 2.80kg)(9.8 m
F  m a

F T1
s
 0.700 m
2
s2
)  29 N
2

F
F m g F ma
T2
T1
m1 g
2
T1
2
F  m ( g  a)  F
T2
Module 10 - 8
1
2
FT 2  (3.50kg)(9.8 m
FT 2  66 N
T1
s
2
 0.70 m
s
2
)  29 N